Lovely Equation

$x={x}+[x]$

Thus, the equation becomes $3\{x\}=2[x]$

Now, $0\leq3\{x\}<3\Rightarrow 0\leq2[x]<3\Rightarrow 0\leq[x]<1.5$

Since $[x]$ is an integer, it can take up values $0,1$

Plugging into the equation, we get $x=0,\frac{5}{3}$

Therefore, $m+n=5+3=\boxed{8}$

${x} = x - [x]$

$4(x -[x]) = x + [x]$

$4x - 4[x] = x + [x]$

$3x = 5[x]$

$\frac{x}{[x]} = \frac{5}{3}$

Now, [x] will always be an integer, so it has to be equal to 1 or 5y

Now, for $[x] =1, x= \frac{5}{3}$

For $[x] = 5y, x = \frac{25y}{3}$ . But this is never possible, as we can easily see.

Let x = I + f, where 0 < f < 1. Putting in the given equation we get 3f = 2I. Plugging integer values to I only I = 1 gives a valid f = 2/3. So the only x = 1 + (2/3) = 5/3.

write the equation as

=> 5 {x} = 2 x

plot the graphs of y=2*x and y=5 * {x}

we find out form the graphs that y=2* x cuts the graph y= 5* {x} at two points;

one being (0,0) and the other being somewhere between x=1 and x=2.

Thus we are now confirmed about two solutions.

To determine the 2nd solution; we write the equation as

=> 3/2 * {x} = [x];

as the 2nd solution lies between x=1 and x=2; [x] has to be equal to 1

therefore {x}=2/3;

thus x= [x] + {x} = 1+ 2/3 = 5/3 =m/n......................[the other solution being 0]

hence m+n=8